package com.potato.study.leetcode.p0455;

import java.util.Arrays;

/**
 * 
 * @author liuzhao11
 * 
 *   455. Assign Cookiesrn
 * 
 *      Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. If sj >= gi, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

Note:
You may assume the greed factor is always positive. 
You cannot assign more than one cookie to one child.

Example 1:
Input: [1,2,3], [1,1]

Output: 1

Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.
Example 2:
Input: [1,2], [1,2,3]

Output: 2

Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.
 * 
 *         思路： 
 *          给每个孩子分小饼干 每个孩子只能分一块 如果达到预期 他们会很满意 求最大的满意数
 *          
 *          排序 greed 和 饼干数  开始遍历 两个数组  i j
 *          while i < g.len j < s.len 
	 *          若 g i  《=     s j 满意 ++ i++ j ++
	 *          若gi 》 sj  j++；
 *          
 *          
 * 			
 * 				
 */	
public class Solution {
	
	public int findContentChildren(int[] g, int[] s) {
		if(null == g || null == s) {
			return 0;
		}
        Arrays.sort(g);
        Arrays.sort(s);
        int i = 0;
        int j = 0;
        int count = 0;//满意的人数
        while(i < g.length && j < s.length) {
        	if(g[i] <= s[j]) {
        		count++;
        		i++;
        		j++;
        	} else { // 不满意
        		j++;
        	}
        }
        return count;
    }
	
	
	public static void main(String[] args) {
		Solution solution = new Solution();
//		int[] g = {1,2,3};
//		int[] s = {1,1};
		int[] g = {1,2};
		int[] s = {1,2,3};
		int count = solution.findContentChildren(g, s);
		System.out.println(count);
		
	}
}
